Bài 1:

a. $2^{29}< 5^{29}< 5^{39}$

$\Rightarrow A< B$

b.

$B=(3^1+3^2)+(3^3+3^4)+(3^5+3^6)+...+(3^{2009}+3^{2010})$

$=3(1+3)+3^3(1+3)+3^5(1+3)+...+3^{2009}(1+3)$

$=(1+3)(3+3^3+3^5+...+3^{2009})$

$=4(3+3^3+3^5+...+3^{2009})\vdots 4$

Mặt khác:

$B=(3+3^2+3^3)+(3^4+3^5+3^6)+....+(3^{2008}+3^{2009}+3^{2010})$

$=3(1+3+3^2)+3^4(1+3+3^2)+...+3^{2008}(1+3+3^2)$

$=(1+3+3^2)(3+3^4+....+3^{2008})=13(3+3^4+...+3^{2008})\vdots 13$

Bài 1:
c.

$A=1-3+3^2-3^3+3^4-...+3^{98}-3^{99}+3^{100}$

$3A=3-3^2+3^3-3^4+3^5-...+3^{99}-3^{100}+3^{101}$

$\Rightarrow A+3A=3^{101}+1$
$\Rightarrow 4A=3^{101}+1$

$\Rightarrow A=\frac{3^{101}+1}{4}$

Bài 2:

a. $7\vdots n+1$

$\Rightarrow n+1\in \left\{1; -1; 7; -7\right\}$

$\Rightarrow n\in \left\{0; -2; 6; -8\right\}$

b.

$2n+5\vdots n+1$
$\Rightarrow 2(n+1)+3\vdots n+1$

$\Rightarrow 3\vdots n+1$

$\Rightarrow n+1\in \left\{1; -1; 3; -3\right\}$

$\Rightarrow n\in \left\{0; -2; 2; -4\right\}$

1 

a=(-5;5;-6;6;-7;7;-8;8;-9;9;.......)

2 bằng nhau

1. a thuộ̣c (+-5, +-6, +-7, +-8,..)

2. |a|<|b|

lộn

2 |a|<|b|

a, Có\(\frac{3n+2}{n}=3+\frac{2}{n}\)

Vì \(3\inℤ\)=> Để \(a\inℤ\)thì \(\frac{2}{n}\inℤ\)<=> \(n\in U\left(2\right)=\left\{\pm1;\pm2\right\}\)

b, Có

\(\frac{a+n}{b+n}=1-\frac{b-a}{b+n}\)

\(\frac{a}{b}=1-\frac{b-a}{b}\)

Vì\(b+n\ge b\)=> \(\frac{b-a}{b+n}\le\frac{b-a}{b}\)=> \(1-\frac{b-a}{b+n}\ge1-\frac{b-a}{b}\)=> \(\frac{a+n}{b+n}\ge\frac{a}{b}\)

CTHH: RO

M_RO = 14.2 = 28 (g/mol)

=> M_R = 28-16 = 12 (g/mol)

=> R là C

=> CTHH: CO

a + b - c = -3 (1) ; a - b + c = 11 (2) ; a - b - c = -1

Từ (1) và (2) ta có:

(a + b - c) + (a - b + c) = - 3 + 11 = 8

a + b - c + a - b + c = 8 => 2a = 8 => a = 4.

Từ (1) và (3) ta có:

( a + b - c) + ( a - b - c ) = -4

a + b - c + a - b - c = -4 => 2a - 2c = -4 => 8 - 2c = 4 => c = 6.

Vậy b = 4 - ( 11 - 6 ) = -1

Vậy ba số nguyên a,b,c cần tìm là 4; -1; 6.